--- title: "Short circuit MVA of feeder circuit breaker" subject: "Power System Analysis II" chapter: 3 key: "semester_5/power_system_analysis_ii:feeder-cb-short-circuit-mva" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/feeder-cb-short-circuit-mva site: BCT Engineering --- # Short circuit MVA of feeder circuit breaker Power System Analysis II · Chapter 3: Power System Fault Calculation - Asked in: 2070 Asar, 8 marks; 2070 Chaitra, 8 marks (asked 2 times) - Video: [SHORT CIRCUIT MVA / FAULT MVA WITH CURRENT LIMITING REACTANCE / KTU/ POWER SYSTEM ANALYSIS](https://www.youtube.com/watch?v=SB1tqp0fGYY) — praveen raj - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/feeder-cb-short-circuit-mva - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3 ## Question For the 3 bus power system shown in figure below the generators are rated 100 MVA with transient reactance of 10% each. Both the transformers are 100 MVA with a leakage reactance of 5%. The reactance of each of the line to a base of 100 MVA, 132 KV is 10%. Find the short circuit MVA of circuit breaker in outgoing feeder from bus-3, if a 3-phase fault occurs beyond the CB at point F. Assume pre-fault voltage of bus-3 is 1 pu and pre-fault current through feeder is zero. [Figure: G1 - T1 (11/132 kV; labelled 11/110 kV in the 2070 Asar paper) - bus 1; G2 - T2 (11/132 kV) - bus 2; lines 1-2, 1-3 and 2-3, each j0.1; outgoing feeder from bus 3 through circuit breaker CB to the fault point F] ## Short answer Each source branch 0.15 pu, lines 0.10 pu; symmetrical network gives Xth = 0.125 pu at bus 3. Short-circuit MVA of feeder CB = 800 MVA, fault current = 3.50 kA at 132 kV. ## Answer The short-circuit MVA that the feeder circuit breaker must interrupt is the fault level at bus 3, since the fault is just beyond the breaker and the feeder impedance up to F is negligible. Base: 100 MVA, 11 kV / 132 kV. Pre-fault $V_3 = 1$ pu and pre-fault current is zero, so all source EMFs are 1 pu. #### Per-unit reactances (100 MVA base) - G1, G2: 0.10 pu each; T1, T2: 0.05 pu each, so each source branch is $0.10 + 0.05 = 0.15$ pu. - Lines 1-2, 1-3, 2-3: 0.10 pu each. ```text E=1 E=1 | | j0.15 j0.15 | | bus1 ---- j0.1 ---- bus2 \ / j0.1 j0.1 \ / bus 3 --CB-- F ``` #### Thevenin reactance at bus 3 The network is symmetrical about bus 3: buses 1 and 2 are always at the same potential, so line 1-2 carries no current. Each path is $0.15 + 0.10 = 0.25$ pu, and the two are in parallel: $$ X_{th} = \frac{0.25 \times 0.25}{0.25 + 0.25} = 0.125\ \text{pu} $$ #### Short-circuit MVA of the breaker $$ \begin{aligned} I_f &= \frac{V_3}{X_{th}} = \frac{1}{0.125} = 8\ \text{pu} \\ \text{SC MVA} &= \frac{S_{base}}{X_{th}} = \frac{100}{0.125} = 800\ \text{MVA} \\ I_f &= 8 \times \frac{100 \times 10^6}{\sqrt{3} \times 132 \times 10^3} = 8 \times 437.39 = 3499.1\ \text{A} \end{aligned} $$ **Answer:** the circuit breaker should be rated for a short-circuit capacity of **800 MVA**, with a fault current of about 3.50 kA at 132 kV. (For the 11/110 kV variant the per-unit values and the 800 MVA result are the same; only the current in amperes changes, to 4.20 kA at 110 kV.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is each source branch reactance (generator plus transformer)? A. 0.15 pu B. 0.10 pu C. 0.05 pu D. 0.25 pu Answer: A. Generator 0.10 and transformer 0.05 are in series, giving 0.15 pu. (Key point: Source branch reactance) 2. Why does line 1-2 carry no current? A. The network is symmetrical about bus 3 so buses 1 and 2 are at equal potential B. It has zero impedance C. It is a radial line D. The generators are rated differently Answer: A. Both paths to bus 3 are identical with equal EMFs, so V1 = V2. (Key point: Symmetry simplification) 3. What is X_th at bus 3 and the breaker short-circuit MVA? A. 0.125 pu and 800 MVA B. 0.25 pu and 400 MVA C. 0.15 pu and 667 MVA D. 0.0625 pu and 1600 MVA Answer: A. Two 0.25 pu paths in parallel give 0.125 pu; SC MVA = 100/0.125 = 800. 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